- #1
Dragonfall
- 1,023
- 5
Randomly permute (1,...,n). What is the probability that exactly i points are fixed?
I think it should be
\binom{n}{i}\frac{(n-i-1)!}{n!}
Is it right?
If so, is the expected number of fixed points (I know it's 1):
\sum_{i=0}^{n}i\binom{n}{i}\frac{(n-i-1)!}{n!}
But it doesn't sum to 1, I think
I think it should be
\binom{n}{i}\frac{(n-i-1)!}{n!}
Is it right?
If so, is the expected number of fixed points (I know it's 1):
\sum_{i=0}^{n}i\binom{n}{i}\frac{(n-i-1)!}{n!}
But it doesn't sum to 1, I think
Last edited:
